How to prove that $\frac{1}{1^2}+\frac{1}{2^2}+\dots+\frac{1}{n^2}+\dots=\frac{\pi^2}{6}$ using the spiral right angle triangle method?

Question

I see this formula given below on You tube video of mathologer channel and then I try to find some new method to prove it:

$$\sum_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}6$$

I tried to prove it geometrically like this

Our attempt:

(1) First I tried to convert it in inverse trignometric form like this but that doesn't help much:

(2) In my second attempt I rotate the length of $1/2$ length from $1$ then I rotate $1/3$ length from remaining $1/2$ but that thing doesn't help us.

(3) In my third attempt, I tried to use coordinate geometry but that makes things more complex.

My question:How to prove that that summation of $1/n^2$ where $n$ tends to infinity is equal to $π^2/6$ by using spiral right angle triangle method?

EDIT

NOTE: Sinc the last line segment Whose length tends to Square root of $π^2/6$ but not exactly equal to Square root of $π^2/6$ so it is probably not possible to solved it by using pure geometry. we understand that there must be needs of theory of Limit to prove it . so we will also accept the solution which take the use of both concept means geometry with slight use of calculus.

I am only aware of the approach with Fourier series to determine the value of this sum. Good luck for your method to work ! — Peter, Jun 16 '22 at 07:20
$1+1/2+1/3+...+1/n$ doesn't converge though, so the circumference of your circle will be infinite, won't it? — Suzu Hirose, Jun 16 '22 at 07:36
A thought: your spiral won’t converge to a point, since the outer edges are the harmonic series with infinite length. The radius will converge as you describe though, so the spiral should approach a circle from the inside and keep going round and round (albeit slowly) — Milten, Jun 16 '22 at 07:39
After $n$ triangles you reach $\prod_{k=1}^n(1+i/f(k))$ with $\prod_{k=1}^n(1+1/[f(k)]^2)=\sum_{k=1}^n1/k^2$. You want to prove $\lim_{n\to\infty}|\prod_{k=1}^n(1+i/f(k))|=\pi/\sqrt{6}$. I can't see this being done without solving the Basel problem another way first. — J.G., Jun 16 '22 at 07:42
@Peter, fourier method is already discussed here: https://math.stackexchange.com/q/3412036/999691 — , Jun 16 '22 at 07:53
You could alternate the direction at which you take the 90 degree turn, and then at least the triangle corners converge. — 2'5 9'2, Jun 16 '22 at 07:59
For a collection of (other) possible solutions see also this post. — Dietrich Burde, Jun 16 '22 at 08:42
@2'59'2 , Can you please provide me details information about how to rotate it by 90°? I don't understand your statement — , Jun 16 '22 at 12:30
@J.G. , you mean that it is impossible to solved without using limits or calculus? I was thinking that there could be some possibility to prove this by Euclidean geometry , Trigonometry or some extra construction ... — , Jun 16 '22 at 12:33
@Milten, You mean that Spiral will never stop? But after some more steps it seems like that line whose length is π^2/6 will stop there......, I mean it looks like that line segment will move but observer will feel that the line segment of length π^2/6 will stop — , Jun 16 '22 at 12:37
Whether $r$ converges is a separate question from whether $\theta$ converges, which is why my previous comment was careful to only talk about moduli or square moduli. However, I can't prove your strategy won't work. — J.G., Jun 16 '22 at 13:14
Here is a non-fully-rigorous proof that your spiral will turn an infinite number of turns: $\arctan \frac{(1/(k+1)}{\sqrt{...}} \approx$ $\arctan \frac{1/(k+1)}{\sqrt{\pi^2/6}}\approx \frac{a}{k+1}$ where $a$ is a constant, but the sum of this series is infinite... — Jean Marie, Jun 16 '22 at 13:34
@JeanMarie, see here :https://www.dropbox.com/s/brprzxgahg731yt/1655387350765.jpg?dl=0 ,...., Sinc k approaches to infinity then last angle approaches to zero....may be i am wrong but can you verify it please — , Jun 16 '22 at 13:50
This is not contradictory with what I have written : harmonic series has its general term tending to 0 but its sum is infinite... — Jean Marie, Jun 16 '22 at 13:53
@JeanMarie , yes you are right that it is tending value and tends to zero but I don't understand about how you claim that line segment of √π^2/6 will turn infinite times? ..... , Because each times we are adding square of reciprocal of natural number then last line segment may be seems moving slowly — , Jun 16 '22 at 13:58
Slowly but surely arbitrarily large : (think to angles in radians), you will sooner or later exceed $2 \pi$, (one turn), then exceed $4 \pi$ (two turns), etc. maybe when you have 1 thousand terms or more... — Jean Marie, Jun 16 '22 at 14:03
@JeanMarie , Interesting but hard to imagine and thanks for clarifications — , Jun 16 '22 at 14:14
@JayendraandSankalp What I mean is, start at $(0,0)$ and go to $(1,0)$. Now you turn left 90 degrees and go to $\left(1,\frac12\right)$. But next, turn right. That is, move in a general clockwise direction $1/3$ of a unit, orthogonal to the vector $\left\langle 1,\frac12\right\rangle$. After that, move $1/4$ of a unit but now in a general counter-clockwise direction. Keep alternating the direction. Your radial lengths are still growing in the same way to $\frac{\pi}{\sqrt{6}}$ but the angle of your position will converge now by a version of the alternating series test. — 2'5 9'2, Jun 16 '22 at 21:15
In the thread it seems that you may not be understanding how the radial length could stabilize even though the radius itself never stabilizes. Here is a graph that may help. It is not the graph of the construction you are investigating, but it does demonstrate that same behavior. — 2'5 9'2, Jun 16 '22 at 21:27
@2'59'2, yes the idea was really great on going clockwise and anticlockwise sence then in imagination i think , if we do continuous clockwise and anticlockwise method as discussed by in your above message then we probably observed that line segment having length approaches √π^2/6 seems as oscillating ......, Thanks for the link of your graph as it helps me lot in understanding about how my construction will looks like — , Jun 17 '22 at 06:51
This is graph of the construction till $n=6400$: https://i.stack.imgur.com/gSt8E.png — Ivan Kaznacheyeu, Jun 17 '22 at 14:40
@JayendraandSankalp Here is a picture of the oscillating version with a few points in the sequence present. The outer circle has radius $\pi/\sqrt{6}$. https://imagebin.ca/v/6kzWdKaDOHfn — 2'5 9'2, Jun 17 '22 at 18:03
@FShrike , this same link is already mentioned by Dietrich Burde in above comment — , Jun 17 '22 at 18:03
@2'59'2 , in your construction , we take *clockwise -antickockwise-clockwise-anticlockwise...* Pattern which leads to Oscillating line similarly I was thinking to take lots of combination of Anticlockwise and clockwise sence....., In my construction i take continuously anticlockwise sence which gives spiral and change into Circle ....., Similarly if we take other combination like *anticlock-antuclock, clockwise-clockwise-anticlock-anticlock*... types of combination then it will produce a line which will oscillates slowly than your construction..[continued] — , Jun 17 '22 at 18:23
And Such type of clockwise and anticlockwise construction gives more number of beautiful patterns.... — , Jun 17 '22 at 18:25

score 1 · Answer 1 · answered Jun 19 '22 at 07:48

Since, using bgeneralized harmonic numbers $$\sum_{i=1}^k \frac 1 {i^2}=H_k^{(2)}$$

quoting @Jean Marie "your spiral will turn an infinite number of turns..." $$A_k=\tan ^{-1}\left(\frac{1}{(k+1) \sqrt{H_k^{(2)}}}\right)$$ which, for large $k$ is $$A_k=\frac{\sqrt{6}}{\pi k}\Bigg[1-\frac{\pi ^2-3}{\pi ^2 k}+\frac{27-13 \pi ^2+2 \pi ^4}{2 \pi ^4 k^2}-\frac{-135+90 \pi ^2-22 \pi ^4+2 \pi ^6}{2 \pi ^6 k^3}+O\left(\frac{1}{k^4}\right) \Bigg]$$

Repeating @Ivan Kaznacheyeu's calculations $$\sum_{k=1}^{6400} A_k=6.678$$ while using the truncated series, we obtain $6.607$

In fact, computed exactly $$\sum_{k=1}^{3857} A_k=2\pi +0.0000601086$$

How to prove that $\frac{1}{1^2}+\frac{1}{2^2}+\dots+\frac{1}{n^2}+\dots=\frac{\pi^2}{6}$ using the spiral right angle triangle method?

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